SOLUTION: I have to figure the dimensions of a fence in a rectangular area, using a 1000 sq ft of my back yard for a dog to run and play in using a Quadratic Equation. Did I calculate this

Algebra ->  Quadratic Equations and Parabolas  -> Quadratic Equations Lessons  -> Quadratic Equation Lesson -> SOLUTION: I have to figure the dimensions of a fence in a rectangular area, using a 1000 sq ft of my back yard for a dog to run and play in using a Quadratic Equation. Did I calculate this       Log On


   

Question 474933: I have to figure the dimensions of a fence in a rectangular area, using a 1000 sq ft of my back yard for a dog to run and play in using a Quadratic Equation. Did I calculate this the width correct? Thank you

L = 2W +5
L = length, W = width, P = perimeter, A = area
〖2w〗^2+5w-1000 = 0
A=2, B=5, C=1000
W = (-b±√(b^2-4ac))/2a
W = (-5±√(5^2-4(2)-1000))/(2(2))
W=(-5±√(5^2-4(2)-1000))/(2(2))
W = (-5±√(25-992))/4
W = (-5±√967)/4
W = (-5±√(-5+241.75))/4 = (236.75)/4 = 59.19 OR (-5-241.75)/4 = (-236.75)/4 = -59.19
W = 59.19 yards of fence
Answer by mananth(16946) About Me  (Show Source):
You can put this solution on YOUR website!
width = W
Length = 2W+5
Area = W(2W+5)=1000
2W^2+5w=1000
2W^2+5w-1000=0
Find the roots of the equation by quadratic formula
a= 2 b= 5 c= -1000

b^2-4ac= 25 - 8000
b^2-4ac= 8025
sqrt%28%098025%09%29= 89.58
x=%28-b%2B-sqrt%28b%5E2-4ac%29%29%2F%282a%29
x1=%28-b%2Bsqrt%28b%5E2-4ac%29%29%2F%282a%29
x1=( -5 + 89.58 )/ 4
x1= 21.15
x2=( -5 -89.58 ) / 4
x2= -23.65
Ignore negative value
x = 21.15ft the width
Length = 2W+5=47.29 ft
CHECK
21.15*47.29=1000
m.ananth@hotmail.ca